Three sinusoidal alternating currents of RMS 5, 7.5, and 10 A are having the same frequency of 50 Hz with the phase angle 30, -70, and 45 degrees respectively. Find the average value in amperes. choices: a. 4.5, 6.76, 9.0 b. 5, 7.5, 10 c. 5.56, 8.83, 11.11 d. 6.0, 8.0, 10.0 Refer to previous problem, write the equations for their instantaneous current in ampere values. choices: a. i1 = 7.07 sin(314t + 30), i2 = 10.61 sin(314t + 70), i3 = 14.14 sin(314t + 45) b. i1 = 7.07 sin(314t + 30), i2 = 10.61 sin(314t - 70), i3 = 14.14 sin(314t + 45) c. i1 = 7.07 sin(314t - 30), i2 = 10.61 sin(314t - 70), i3 = 14.14 sin(314t + 45) d. i1 = 7.07 sin(314t - 30), i2 = 10.61 sin(314t + 70), i3 = 14.14 sin(314t + 45)
Three sinusoidal alternating currents of RMS 5, 7.5, and 10 A are having the same frequency of 50 Hz with the phase angle 30, -70, and 45 degrees respectively. Find the average value in amperes. choices: a. 4.5, 6.76, 9.0 b. 5, 7.5, 10 c. 5.56, 8.83, 11.11 d. 6.0, 8.0, 10.0 Refer to previous problem, write the equations for their instantaneous current in ampere values. choices: a. i1 = 7.07 sin(314t + 30), i2 = 10.61 sin(314t + 70), i3 = 14.14 sin(314t + 45) b. i1 = 7.07 sin(314t + 30), i2 = 10.61 sin(314t - 70), i3 = 14.14 sin(314t + 45) c. i1 = 7.07 sin(314t - 30), i2 = 10.61 sin(314t - 70), i3 = 14.14 sin(314t + 45) d. i1 = 7.07 sin(314t - 30), i2 = 10.61 sin(314t + 70), i3 = 14.14 sin(314t + 45)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
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Three sinusoidal alternating currents of RMS 5, 7.5, and 10 A are having the same frequency of 50 Hz with the phase angle 30, -70, and 45 degrees respectively. Find the average value in amperes.
choices:
a. 4.5, 6.76, 9.0
b. 5, 7.5, 10
c. 5.56, 8.83, 11.11
d. 6.0, 8.0, 10.0
Refer to previous problem, write the equations for their instantaneous current in ampere values.
choices:
a. i1 = 7.07 sin(314t + 30), i2 = 10.61 sin(314t + 70), i3 = 14.14 sin(314t + 45)
b. i1 = 7.07 sin(314t + 30), i2 = 10.61 sin(314t - 70), i3 = 14.14 sin(314t + 45)
c. i1 = 7.07 sin(314t - 30), i2 = 10.61 sin(314t - 70), i3 = 14.14 sin(314t + 45)
d. i1 = 7.07 sin(314t - 30), i2 = 10.61 sin(314t + 70), i3 = 14.14 sin(314t + 45)
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